Question

函数imblearn.under_sampling.RandomUnderSampler仅允许我通过dict输入所需的欠采样百分比作为绝对数字，但是绝对数字会干扰（时间序列）交叉验证，而我没有相同级别的少数族裔样本每折一次。（这会产生持续的错误：Originally, there is 11037 samples and 28546 samples are asked.）

是否可以输入相对值，例如0类为80％，1类为20％，等等？

Answer 1

我认为这是一个最小的工作示例。那解决了你的问题。从馆藏进口柜台从sklearn.datasets导入make_classification 从sklearn.model_selection导入KFold 从sklearn.pipeline导入管道从imblearn.under_sampling导入RandomUnderSampler

def classify(datasets, labels, *args):
    kf=KFold(n_splits=3)
    for train_idx, test_idx in kf.split(X):
        print('Original dataset shape {}'.format(Counter(labels[train_idx])))
        train_x, train_y = datasets[train_idx], labels[train_idx]
        test_x, test_y = datasets[test_idx], labels[test_idx]
        ratio_dict = {}
        for k,v in enumerate(args):
            ratio_dict.update({k:int ((v/ 100) * Counter(train_y)[k])})
        print(ratio_dict)
        rus = RandomUnderSampler(random_state=42, ratio=ratio_dict)
        X_res, y_res = rus.fit_sample(X, y)
        print('Resampled dataset shape {}'.format(Counter(y_res)))


X, y = make_classification(n_classes=2, class_sep=2,
    weights=[0.1, 0.9], n_informative=3, n_redundant=1, flip_y=0,
    n_features=20, n_clusters_per_class=1, n_samples=1000, random_state=10)

rus = classify(X, y, 10, 20)

我很想看看是否有人可以使用sklearn的Pipeline Framework来实现这一点。

具有相对而非绝对比率的随机欠采样

1 个答案: